Conduct practical investigations to analyse the reversibility of chemical reactions (Chapter 2). 

• Reactions can be classified as either reversible or irreversible. 

• Many reactions that appear irreversible are indeed reversible under extreme conditions. 

• Dynamic equilibrium – a chemical equilibrium where the reaction is proceeding in both directions at the same rate, with no net change. 

• Equilibrium is not achieved in an open system.  

• Reversible reaction – a chemical reaction where the reactants form products that, in turn, react together to give the reactants back. 

• Reversible relations in a closed system eventually reach a point where rates of forward and reverse reactions are equal with no further change – equilibrium has been achieved. 

• Some reversible processes – evaporation or condensation of water, saturated sugar solution. 

• An example of an everyday irreversible reaction is baking a cake – reactions occur changing the chemical structures of the proteins in flower and egg, as well as reactions between the proteins and sugars. 

• Le Chatelier’s Principle – system at equilibrium will resist stress applied and find a new equilibrium position. 

• Explaining reversibility: 

• When particles collide, the energy associated with the collisions can break bonds in reacting particles, allowing them to rearrange and form new products. 

• Once products form, if the product particles collide with energy equal to or greater than the activation energy of the reverse reaction, the original reactants can be re-formed.  

• Prior to any stress applied, the reaction is occurring in both directions (macroscopic changes). 

• Cobalt chloride reaction: CoCl2 (s) + 6H2O (l) ßà CoCl2 x 6H2O (s). When a stress is applied we move position of the equilibrium and either reactant/product becomes more dominant. A new equilibrium is established, but the reaction is still easily reversed. 

• Paper impregnated with blue dehydrated cobalt (II) chloride can be used to detect water, turning pink as hydrated cobalt (II) chloride is formed (making it useful as a moisture indicator in weather instruments). 

• This reaction is reversible as when heated the water evaporates, regaining its blue colour. 

• Iron (III) nitrate and potassium thiocyanate: Fe (3+) (aq) + SCN- ßà FeSCN (2+) (aq). The iron is a pale-yellow colour and potassium colourless, forming a red-coloured ion. 

• If temperature is increased, reaction will move to direction that reduce overall temp (look at enthalpy).  

• Static equilibrium – a reaction where there is no exchange between reactants and products and we consider reaction to have gone to completion and be irreversible (rates of forward and reverse reactions are zero).  

• Combustion of magnesium: 2Mg (s) + O2 (g) à MgO (s). Combustion reactions release great heat and require great heat to reverse. Hence, they are considered irreversible. 

• Combustion of steel wool: 4Fe (s) + 3O2 (g) à 2Fe2O3 (s). Combustion of steel wool produces iron oxide (a very low energy state). To reverse this, great energy is required (smelting of haematite). This is considered irreversible under most laboratory conditions.  

• Combustion involves burning an organic compound (hydrocarbon) with oxygen to form CO2 and H2O. Because water and carbon dioxide are stable, they don’t react to form the reactants. 

• If there is not enough oxygen to supply the combustion, either CO or C is present on the reactants. 

• Evaporation or condensation of water: H2O (l) ßà H2O (g). Evaporation of water from rivers, cloud formation and eventual rain is a physical change. This shows water can cycle between different states because each phase is reversible. 

Model static and dynamic equilibrium and analyse the differences between open and closed systems. 

• Open systems – open to exchange both matter and energy (e.g. heat) with the external environment. 

• Closed systems – open to exchange only energy (e.g. heat) with the external environment. 

• Isolated systems – not open to exchange either energy or matter with the external environment (a perfect isolated system is almost impossible). 

• Steps of equilibrium – (1) Reactants are added. (2) Reaction commences in the forward direction. (3) Reaction moves in reverse direction to counteract increase in conc. of products. (4) Dynamic equilibrium established. Note – concentrations are not necessarily equal, centre isn’t in centre of 2 reactions. 

• Steady state – system not in equilibrium where conditions are stable and there is no net change over time. It is a physical system where matter and energy are leaving system at a constant rate so conditions have become stable. 

• Dynamic state of equilibrium:  

• At a molecular level, bonds are constantly being broken and new bonds are formed as the reactants and products continue to be converted from one to another. 

• Decomposition of N2O4: N2O4 (g) ßà 2NO2 (g). A vessel injected with pure N2O4 begins colourless but after seconds starts turning into a dark brown gas. When equilibrium is reached, colour does not change. 

• Static equilibrium: 

• Static equilibrium occurs when the rates of the forward/reverse reactions are both almost zero. 

• In these reactions, there is no further conversion of reactants to products or vice-versa. 

• Conversion of graphite to diamond is a static equilibrium example: C (graphite) ßà C (diamond). 

• In theory, diamond has to be heated above 2000 degrees due to the high reverse activation energy. 

• Under normal conditions, this would take billions of years as the reactions rates are virtually zero. 

Analyse examples of non-equilibrium systems in terms of the effect of entropy and enthalpy. 

• A non-equilibrium system is a reaction that goes to completion. 

• This will happen when the reaction is spontaneously favourable (exergonic). 

• Entropy - a measure of molecular disorder, showing the possible ways energy can be distributed in a system. The more spread out, the higher the entropy and the more likely a reaction will occur spontaneously. This is measured in J/K. 

• Spontaneous reaction – a reaction which is thermodynamically capable of occurring without external energy input to begin it. 

• Exergonic – reactions that release energy and occur spontaneously. 

• Endergonic – reactions that are non-spontaneous and require input of energy to continue. 

• Non-equilibrium reactions:  

• Some systems are effectively irreversible and never reach an equilibrium state. 

• Combustion and photosynthesis are examples of non-equilibrium systems. 

• Combustion: 

• Combustion reactions are irreversible, exothermic and spontaneous non-equilibrium systems. 

• Some examples include the combustion of octane and methane to produce CO2 (g) + H2O (g). 

• The products formed are stable and do not re-form octane/methane and oxygen. 

• The octane reaction involves increase in entropy as no. of gas molecules increases from 13.5 to 17. 

• It is a spontaneous (-G) system and does not need continuing energy supply to occur. 

• Photosynthesis:  

• Photosynthesis is an endothermic, non-equilibrium chemical process by plants and algae to convert CO2 and H2O into oxygen and glucose – 6CO2 (g) + 6H2O (l) à C6H12O6 (aq) + 6O2 (g). 

• During this reaction, heat energy is also produced and released to the surroundings (increasing the entropy of the surroundings). 

• Photosynthesis involves a decrease in entropy as the 12 reactant particles become 7. 

• It is a non-spontaneous (+G) system requiring continuous energy supply for the reaction to occur. 

• The photosynthesis process can occur as it is coupled with spontaneous reactions, with the energy released by the spontaneous reactions used during the many small steps of photosynthesis. 

Investigate the relationship between collision theory and reaction rate in order to analyse chemical equilibrium reactions. 

• Rate of reaction – the speed at which reactants are converted into products dependent on the frequency of collision of chemical species. 

• Increasing the energy or chance of collisions in the correct orientation will increase RoR. 

• Reactions occur when particles collide with enough kinetic energy and the correct orientation to break bonds and form an activation complex for new substances to be formed. 

• If activation energy is low enough in both directions, the reaction can occur in equilibrium.  

• Due to LCP, methods of changing a RoR (pressure, concentration, temp) will affect equilibrium position. 

• The fact that many reactions don’t proceed to completion has consequences for chemical industries.  

• Large amounts of unreacted starting materials are wasteful and costly – particularly when profitability depends on the yield of the reaction (extent of conversion of reactants into products). 

• Equilibrium and collision theory: 

• Consider the Haber Process (2N2 + 3H2 ßà 2NH3). 

• Reversible reactions in closed system eventually reach point where rate of forward = reverse reac. 

• When this situation is reached, ammonia is formed at the same rate as it is breaking down. 

• A concentration versus time graph shows the changes in concentrations of chemicals with time. 

• On these graphs, equilibrium is reached when there is no longer a change in any of these conc. 

• Extent of reaction:  

• Extent of reaction describes how much product is formed when the system reaches equilibrium. 

• Rate of reaction is a measure of the change in conc. of the reactants and products with time and is not directly related to the extent of reaction. 

Deduce the equilibrium expression (in terms of Keq) for homogenous reactions occurring in solution (Chapter 3). 

• Homogenous equilibrium – an equilibrium where all substances are in the same phase e.g. all are gases. 

• Heterogenous equilibrium – an equilibrium where the reactants and products are in different phases. 

• For heterogenous reactions, the concentration of a pure solid or liquid is one – because concentrations do not depend on how much of a pure substance is present. 

• This is the reasoning why a solid isn’t part of the equilibrium expression – it is a constant. 

• Equilibrium expression – a numerical ratio showing concentration of products raised to coefficients to concentration of reactants raised to coefficients. It is a temperature dependent thermodynamic quantity. 

• Although Keq is temperature dependent, it is not affected by pressure or catalysts. 

• As Keq is reliant on temperature, a different temperature would result in a different constant value. 

• Keq depends on direction the equation is written – due to location of the products and reactants in the ratio. 

• Meaning of the value of an equilibrium constant: 

• Value is based on the equilibrium concentrations of the products divided by the equilibrium concentrations of the reactants. 

• Hence, it indicates the extent of reaction at equilibrium – how far forward reaction endures before equilibrium is established and the equilibrium yield (amount of products at equilibrium). 

Perform calculations to find the value of Keq and concentrations of substances within an equilibrium system and use these values to make predictions on the direction in which a reaction may proceed. 

• Equilibrium constant expression: 

• This is a particular ratio at a particular temperature when the system has reached equilibrium. 

• All species are included in the expression where their concentrations can vary – this means we can include concentrations of gases and aqueous solutions, but no solids or pure liquids. 

• Units in the expression calculation are mol/L but Keq has no units. 

• The value of Keq predicts the equilibrium position of the reaction. 

• The size of Keq also indicated the proportions of reactions and products in the equilibrium mixture. 

• If one equation is reverse of another (e.g. 2NO2 ßà N2O4), equilibrium constants are reciprocals. 

• If K > 1, the equilibrium position lies to the right and products are favoured. 

• If K < 1, the equilibrium position lies to the left and reactants are favoured. 

• Because the equilibrium expression is a ratio, we can not get a value of K = 0. 

• Magnitude of Keq: 

• If Keq is approximately 1 (0.0001-10000) – roughly equivalent concentrations of reactants and products (equilibrium position will be in the middle). 

• If Keq is > 1 (10000+) – largely mostly products with a reaction which can be said has gone to completion (equilibrium position is far to the right).  

• If Keq < 1 (0.0001 and smaller) – mostly reactants with a reaction that has occurred to a very small extent (equilibrium position is far to the left). 

• Data needed to calculate Keq – (a) equation, (b) Keq formula, (c) equilibrium concentrations for all species or initial concentration for all species and equilibrium concentration of one species, (d) ICE box. 

• Reaction quotient (also known as concentration fraction): 

• Definition – ratio of products to reactants in any chemical system. 

• It is calculated to determine if a system is currently at equilibrium and the position of equilibrium/favoured reaction. 

• It is calculated in the same manner and using the same expression as Keq, but with measured values. 

• While reaction quotient can be calculated at any time, it only gives a constant value at equilibrium. 

• If Q = Keq – the system is at equilibrium. 

• If Q > Keq – the reverse reaction will be favoured to form reactants (shifts left). 

• If Q < Keq – the forward reaction will be favoured to form products (shifts right). 

• By comparing the values of Q and Keq at a particular temp, it is possible to predict the direction in which a reaction will proceed to reach equilibrium. 

Qualitatively analyse the effect of temperature on the value of Keq. 

• Temperature and Keq – exothermic reactions: 

• Example – Fe(3+) + NH4SCN ßà Fe(SCN)2+ + NH4+ + Heat. 

• If heat is added, the concentration of reactants increases by using up the products. 

• Increase in temperature – decreases Keq and shifts position to the left. 

• Decrease in temperature – increases Keq and shifts position to the right. 

• Temperature and Keq – endothermic reactions: 

• Example – Co(H2O)6 (2+) + 4Cl- + Heat ßà CoCl4 (2-) + 6H2O. 

• If heat is added, the concentration of products increases by using up the reactants. 

• Increase in temperature – increases Keq and shift position to the right. 

• Decrease in temperature – decrease Keq and shift position to the left. 

Conduct an investigation to determine Keq of a chemical equilibrium system. 

• Equation – Fe (3+) (aq) + SCN- ßà FeSCN (2+) (aq) where Keq =  9 x 100 at 25 degrees Celsius.  

• A substantial amount of the product is present in an equilibrium mixture, so the mixture is bright red. 

• This reaction is used in theatres to make fake blood – colourless SCN- is painted onto an actor’s hand, which when interacts with a knife coated in Fe (3+), produces a red liquid.  

Explore the use of Keq for different types of chemical reactions, including but not limited to – dissociation of ionic solutions and dissociation of acids and bases. 

• Dissociation – process in which chemical species split into smaller particles such as ions, usually in a reversible manner.  

• Solubility equilibria – dynamic equilibrium exists when a chemical compound in the solid state is in chemical equilibrium with a solution of that compound. 

• Ionic solids dissociate into ions when they dissolve in water. 

• This equilibrium can involve precipitation reactions or dissociation. 

• Species which exist in solubility equilibria have the solid on the reactants side of the chemical equation. 

• Considering this is a heterogeneous equilibrium, we leave the solid out of the equation. 

• The higher the Ksp value, the more soluble the substance. 

• Ksp only represents solutions at equilibria – otherwise Q (called the ionic product) is used. 

Investigate the effects of temperature, concentration, volume and/or pressure on a system at equilibrium and explain how Le Chatelier’s principle can be used to predict such effects (Chapter 4). 

• Effects of changes are significant to the chemical industry, as conditions must be carefully selected to ensure that optimum yield of products are obtained within a reasonable timeframe. 

• The relative amounts of reactants and products at equilibrium is called the position of equilibrium. 

• At equilibrium, the rates of the forward and reverse reactions are equal. 

• Factors influencing equilibrium – adding or removing a product or reactant, changing the pressure by changing volume (gases), dilution (solutions) and changing the temperature. 

• LCP – when a system at equilibrium experiences changes, it will move to minimise changes based on the thermodynamic properties (endo/exothermic), conc. of reactants/products and pressure/volume changes. 

• By understanding LCP, you can predict the effect of changes on equilibrium systems. 

• Note that the change is ‘partially opposed,’ and will never return to the initial equilibrium position. 

• Note that only temperature changes the value of Keq for an equilibrium system. 

• Effect of temperature: 

• If heat is added to the system, it will move to favour the endothermic reaction. This will decrease the overall temperature and minimise the initial increase. 

• If heat is subtracted, it will move to favour the exothermic reaction. This will increase overall temperature and minimise initial decrease. 

• Effect of pressure: 

• The pressure of a gas is inversely proportional to the volume of its container.  

• Hence, the pressure of gases in an equilibrium mixture can be changed by changing the volume. 

• Increasing pressure (decrease volume) causes system to minimise change by favouring reaction which produces least gaseous molecules. 

• Decreasing pressure (increases volume) causes system to minimise change by favouring reaction which produces most gaseous molecules. 

• Effect of adding an inert gas: 

• Total pressure of an equilibrium mixture of gases can be changed by adding an inert gas – He, Ne, Ar. 

• As there is no change in concentrations, there is no effect on the position of equilibrium.  

• Collision theory shows any collisions with inert gas molecules wont produce reaction – no EP change. 

• Effect of concentration: 

• Concentration can be altered by removing or adding these reagents. 

• If conc. of reactants is increased, system will favour forward reaction to use up excess reactants. 

• If conc. of products is increased, system will favour reverse reaction to use up excess products. 

• Effect of dilution: 

• The focus for an equilibrium in solution is the number of particles per volume of solution. 

• Dilution by adding water reduces the number of particles per volume – results in a shift towards the side that produces the greater number of dissolved particles. 

• Example – Fe (3+) (aq) + SCN- ßà FeSCN (2+) (aq) – addition of water would shift to the left as there are two dissolved particles compared to one. 

• At the time of dilution, there is an instantaneous decrease in the concentration of all species. 

• Predicting effects using the equilibrium law: 

• The equilibrium law expression Keq = products/reactants and the reaction quotient can also be used. 

• If the Q value is < Keq value at equilibrium, we can predict the changes (or vice-versa). 

Explain the overall observations about equilibrium in terms of the collision theory. 

• Collision theory is used to explain the different rates of chemical reactions.  

• Definition – particles only react successfully if they collide under specific conditions.  

• It states that for a reaction to occur, the reactant particles must collide – a larger number of successful collisions in a specified time period results in a faster reaction rate. 

• Pressure: 

• When considering changes in pressure, we can predict changes by looking at molecules. 

• Increased pressure – gas molecules are closer, and collision become more frequent. 

• This means the rate of the reaction involving the greater amount of molecules become greater than the one with a smaller number. 

• As more product is formed, the rate of the opposite reaction increases to balance it out. 

• Pressure changes do not affect the equilibrium position of systems in the liquid or solid phases – particles in these systems are too tightly packed for a pressure change to effect volume. 

• This means there is negligible change in conc. of species involved and no effect on the conc. fraction. 

• What if there is an equal amount of particles on both sides? 

• Equal numbers do not change the position of equilibrium. 

• According to LCP, it doesn’t matter which way the system shifts, the number of particles in the container will remain constant. 

• Hence, the system is unable to oppose the change applied and there is no net reaction. 

• In regard to collision theory, the volume decrease causes the rates of the forward and back reactions to be increased equally. 

Examine how activation energy and heat of reaction affect the position of equilibrium. 

• Changing the temp of an equilibrium in a closed system affect both the position of equilibrium and the value of the equilibrium constant. 

• Activation energy (Ea) – the kinetic energy required for a reaction to take place. 

• Ea needs to be achieved for any reaction to occur – in an equilibrium reactions, the system needs to have enough kinetic energy for the reaction to occur in both directions. 

• The heat of reaction describes whether energy is released or absorbed as the reaction proceeds. 

• Being a thermodynamic property, movement of energy into and out of system affects equilibrium position. 

• 2NO2 (g) ßà N2O4 (g) + energy: 

• This shows the exothermic reaction of a brown gas to a colourless gas. 

• Note that with a change in temp, there is no instantaneous change in concentration. 

• As the product/reactant have different colours, the change in equilibrium can be monitored visually. 

• Increase in temp means a higher conc of NO2, less conc of N2O4. 

• Decrease in temp means a lower conc of NO2, higher conc of N2O4. 

• Exothermic reactions: 

• Increased temp results in a net reverse reaction and a decrease in Keq. 

• Decreased temp results in a net forward reaction and an increase in Keq. 

• Endothermic reactions: 

• Increased temp results in a net forward reaction and an increase in Keq. 

• Decreased temp results in a net reverse reaction and a decrease in Keq. 

• Temperature and collision theory: 

• An increase in temperature causes molecules to move faster – creating more frequent and energetic collisions (larger number of molecules have the necessary energy to overcome the Ea barrier). 

• At the higher temp, rates of both forward/reverse reactions increase. 

• Because the Ea for the endothermic reaction is greater than that of the exothermic reaction, the increased energy will mean there will be a greater proportion of molecules to overcome Ea barrier for the endothermic reaction. 

• The rate of the endothermic reaction will increase more than that of the exothermic. 

• Effect of a catalyst on equilibrium: 

• A catalyst lowers the activation energy of the forward and reverse reactions by the same amount. 

• This lower Ea causes an increase in the number of effective collisions. 

• Hence, there is an increase in the rate of both forward and reverse reactions – because more particles have energies greater than the Ea barrier for the reaction. 

• Due to the equal increases, concentrations of products and reactants don’t change. 

• A catalyst will increase rate at which reaction proceeds towards equilibrium, but not change Keq/ EP. 

• Co(H2O)6 (2+) (aq) + 4Cl- (aq) ßà CoCl4 (2-) (aq) + 6H2O (l): 

• When cobalt (II) chloride is dissolved in water, it forms a pink solution due to the Co(H2O)6 (2+). 

• If cobalt (II) chloride is dissolved in acid, the Co(H2O)6 (2+) ions form an equilibrium with Cl- ions. 

• The forward reaction is endothermic – if heated, the position moves to the right and the colour progressively becomes bluer. 

• Higher temperatures: 

• At higher temperatures, the Ea of both directions is easily achieved. 

• The endothermic reaction has a higher Ea. 

• When the system loses energy, the equilibrium position will move towards exothermic position. 

Describe and analyse the processes involved in the dissolution of ionic compounds in water (C5). 

• Key definitions:  

• Aquo-cation – a hydrated cation surrounded by electrostatically attracted H2O. 

• Aquo-anion – a hydrated anion surrounded by electrostatically attracted H2O. 

• Ligand – an ion or molecule that binds to a central metal atom to form a complex. 

• Ion-dipole bond – an electrostatic bond that occurs when sufficient differences in electronegativity appear between dipoles and charged ions. 

• Solvent – the major component of a solution. 

• Solute – the minor component of a solution. 

• Process of dissolution: 

• Ions are pulled away from the solid crystal lattice by the electrostatic charge or ion-dipole attraction between the water and the ions. 

• This creates a water ligand as they form complexes around the ions, preventing them from reforming the crystal lattice. 

• Aquo-cations and aquo-anions are formed. 

• When no more are able to be formed and the dissolution enters dynamic equilibria, the solution is said to be saturated. 

• When two aqueous reactants are combined in a reaction vessel, the dissolved reactant particles mix freely. 

• This increases chances of the reactants coming into contact – increased particle movement makes the interaction between reactants more effective than if the reactants were mixed as solids. 

• Dissolution of ionic substances in water: 

• Water molecules are polar – oxygen atom has a slight -ve charge and the hydrogens slight +ve. 

• When an ionic substance is added to water, the hydrogen atoms are attracted to the negatively charged ions and the oxygen atoms to the positively charged ions. 

• This attraction is known as ion-dipole attraction. 

• When these ion-dipole attractions are strong enough, the ions are removed from the ionic lattice and the ion is surrounded by water molecules – it is now hydrated. 

• Dissolution of an ionic compound in water: 

• Example – sports drinks are used by athletes to replace the electrolytes lost in sweat – the electrolytes are dissolved ionic substances such as NaCl and potassium phosphate. 

• Process of separating +ve and -ve ions from a solid ionic compound to form hydrated ions when dissolved in water is called dissociation. 

• Example – process of NaCl dissolving in water includes (a) ionic bonds in NaCl are broken, (b) hydrogen bonds between water molecules are broken, (c) ion-dipole attractions. 

• Dissociation of ionic compounds is simply freeing ions from the lattice, so they move freely throughout the solution. 

• Ionic substances dissolve by dissociation – ion-dipole attractions are formed between ions and water molecules. 

• Within a biological context: 

• During digestion, food is broken down into ions and small molecules as dissolved in water in the intestine then absorbed into the blood and lymph to circulate around the body 

• Blood is 80-90% water and it transports a range of ionic and molecular substances (e.g. vitamins, minerals, metabolic wastes) around the body through the walls of cells and capillaries in aqueous solution 

• Within a domestic and industrial context: 

  In maintaining hygiene, cleaning by washing in water relies on the ability of the water to dissolve a large range of substances. Soaps and detergents are used to increase cleaning ability by allowing non-polar substances to dissolve in water 

• In agriculture, essential minerals for plant growth (nitrogen and phosphorous), enter the plant from the soil in the form of nitrates and phosphates. These dissolve in water, allowing them to pass through roots 

• Solubility is dependent on: 

• Ability to attract/separate ions from the lattice (strength of ionic bonding vs. ion-dipole interactions) – this is an endothermic process. 

• Ability to form hydrated ions (strength of the ion-dipole interaction) – this is an exothermic process. 

• The more this process is overall exothermic, the more salt will dissolve. 

Investigate the use of solubility equilibria by Aboriginal and Torres Strait Islander Peoples when removing toxicity from foods. 

• Cycad seeds are toxic if untreated due to cycasin which is a neurotoxin (destructive to nerve tissue). 

• Its solubility is the key to the use of the Cycad – cycasin is soluble at 56.5g/L. 

• By placing the Cycad in flowing water, the cycasin will continue to dissolve out over a long period of time. 

• If seeds were placed in non-flowing water, an equilibrium would be established, and water would need to be drained and replaced every time 56.5g/L had been leached out. 

• Cycad fruit are either pulped or left whole for leaching – pulping creates a larger surface area meaning the cycasin will each away faster. 

• Leaching – drain away soluble salts from a material by the action of percolating liquid or flowing liquid. 

• The bitter yam contains oxalates that would be toxic if consumed, with a detoxification ceremony occurring.  

• The yams are also placed in running water to undergo leaching, and then roasted in an earth oven. 

Conduct an investigation to determine solubility rules and predict and analyse the composition of substances when two ionic solutions are mixed. 

• Saturated solutions: 

• Saturated solution – a solution that contains the maximum amount of solute that can be dissolved at a particular temp. 

• Unsaturated solution – a solution that contains less than the maximum amount of solute that can be dissolved at a particular temp. 

• Insoluble compounds are called sparingly soluble. 

• Definition of solubility: 

• Solubility refers to the maximum amount of a solute than can be dissolved in a given quantity of solvent at a certain temperature. 

• Soluble substance – more than 0.1 mol of substance will dissolve in 1 L of water. 

• Insoluble substance – less than 0.01 mol of substance will dissolve in 1 L of water. 

• Slightly soluble substance – 0.01-0.1 mol of substance will dissolve in 1 L of water. 

• A precipitate is an insoluble solid formed from two ionic solutions. 

• Solubility rules: 

• SNAAP – Sodium, Nitrate, Ammonium, Acetate and Potassium ions will never form precipitates. 

• Insoluble ionic compounds: 

• Solubility is a measure of how much of the substance will dissolve in a given amount of solvent. 

• Example – NaCl is very soluble in water, making it useful for use in saline drips.  

• Insoluble compounds do not dissolve as the energy required to separate the ions from the lattice is greater than the energy released when the ions are hydrated. 

• A substance is rarely ever completely soluble/insoluble – a solubility scale is more accurate. 

• Potassium chloride and silver nitrate – KCl (aq) + AgNO3 (aq) à KNO3 (aq) + AgCl (s). 

• Potassium iodide and lead (II) nitrate – KI (aq) + Pb(NO3)2 (aq) à 2KNO3 (aq) + PbI2 (s). 

• Sodium sulphate and barium nitrate – Na2SO4 (aq) + BaNO3 (aq) à 2NaNO3 (aq) + BaSO4 (s) 

Derive equilibrium expressions for saturated solutions in terms of Ksp and calculate the solubility of an ionic substance from its Ksp value. 

• Ksp is essentially the Keq for a solid dissolving in water – except all reactants are solids and thus ignored. 

• Ksp indicates the degree to which a compound dissociates in water – higher constant, more soluble. 

• Saturated solution exist in a dynamic equilibrium and these temperature dependent concentrations are the values that the reference value for Ksp are derived from. 

• Saturated solution – a solution that has dissolved the maximum amount of solute for that temperature and now exists in a dynamic equilibrium.  

• Molar solubility – the number of moles of a substance that can be dissolved per litre of solution before the solution becomes saturated. 

• The solubility of ionic compounds in water depends on the differences in the energy required to separate the ions from the lattice, and the energy released when the ions are hydrated.  

• Saturated solutions: 

• Unsaturated solutions – solutions that contain less than the maximum amount of solute that can be dissolved at a specific temperature. 

• Sparingly soluble ionic compounds dissolve to a very small extent – very small amounts of these salts are needed to form a saturated solution e.g. silver chloride. 

• At an atomic level, the AgCl lattice is constantly changing, with ions leaving it and returning at the same rate – it is at equilibrium according to AgCl (s) ßà Ag+ + Cl-. 

• When a soluble substance dissolves completely, only the forward arrow is used. 

• For sparingly soluble salts, double arrow shows the rate of dissolution = rate of association 

• Solubility product: 

• In writing the equilibrium expression for a sparingly soluble salt, only aqueous ions are included. 

• Solubility product shows the product of the concentration of ions in a saturated solution of a sparingly soluble salt. 

• The ionic product = Ksp only when a solution is saturated. 

• Ksp cannot always be used to compare the relative solubility of salts as different salts produce different numbers of ions in solution. 

• Ksp can only be used to compare the solubility of salts that produce same number of ions. 

• Example – AgCl and CaSO4 both produce two ions, so their Ksp can be compared (smaller Ksp value shows less solubility). 

Predict the formation of a precipitate given the standard reference values for Ksp. 

• The solubility constant can be used to calculate the concentration of ions in a saturated solution, calculate the solubility of an ionic compound or predict the formation of a precipitate. 

• Predicting precipitation reactions: 

• Ionic product < Ksp – unsaturated and no precipitate. 

• Ionic product = Ksp – saturated and precipitate formed. 

• Ionic product > Ksp – supersaturated and precipitate formed. 

• Common ion effect: 

• Common ion effect refers to when a salt solution is added to an existing solution with a common ion. 

• This greatly decreases the solubility of ionic compounds, meaning that they are far less soluble in a solution with a common ion compared to water. 

• Example – if you add sodium chloride solution to saturated lead chloride solution, the additional Cl- ions will drive the equilibrium position to favour the lead chloride solid, decreasing its solubility.